Power Series Expansion for Reciprocal of Cube Root of 1 + x
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Theorem
Let $x \in \R$ such that $-1 < x \le 1$.
Then:
- $\dfrac 1 {\sqrt [3] {1 + x} } = 1 - \dfrac 1 3 x + \dfrac {1 \times 4} {3 \times 6} x^2 - \dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots$
Proof
\(\ds \frac 1 {\sqrt [3] {1 + x} }\) | \(=\) | \(\ds \paren {1 + x}^{-\frac 1 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-\frac 1 3}^{\underline k} } {k!} x^k\) | General Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-\frac 1 3} - j} } {k!} x^k\) | Definition of Falling Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {3 j - 1} } {3^k k!} x^k\) | Translation of Index Variable of Product and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \frac 1 3 x + \frac {1 \times 4} {3 \times 6} x^2 - \frac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots\) | extracting the juice |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 20$: Binomial Series: $20.13$